A circle has a sector with area $\dfrac{40}{3}\pi$ and central angle $\dfrac{5}{12}\pi$ radian. What is the area of the circle? ${64\pi}$ $\color{#9D38BD}{\dfrac{5}{12}\pi}$ ${\dfrac{40}{3}\pi}$
Solution: The ratio between the sector's central angle $\theta$ and $2 \pi$ radians is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{2 \pi} = \dfrac{A_s}{A_c}$ $\dfrac{5}{12}\pi \div 2 \pi = \dfrac{40}{3}\pi \div A_c$ $\dfrac{5}{24} = \dfrac{40}{3}\pi \div A_c$ $A_c \times \dfrac{5}{24} = \dfrac{40}{3}\pi$ $A_c = \dfrac{40}{3}\pi \times \dfrac{24}{5}$ $A_c = 64\pi$